In $\Delta ABC$, $\angle C = 36^{\circ}$ and $\angle B = 96^{\circ}$. What is $\angle MNC$?

Question

In $\Delta ABC$, $\angle C = 36^{\circ}$ and $\angle B = 96^{\circ}$. $N$ lies on $AC$ such that $AN = NE$, where $E$ is defined such that $CE = AB$. $M$ is the midpoint of $BC$. What is $\angle MNC$?

I got this question from a friend on Discord who also was stuck. So I drew the diagram on GeoGebra and the answer turns out to be $24^{\circ}$ (https://www.geogebra.org/geometry/cmdxcdds). The condition $CE = AB$ really stood out to me, so I defined point $D$ such that $AB = BD$, and it turns out that $BD \perp NM$. I have no idea how to prove this however, and I haven't had any success with using other theorems such as the angle in the circle being twice the angle on the circumference.

I would preferably like a solution involving only Euclidean geometry, but a solution with trigonometry is fine.

Answer 1

Let $R$ be the midpoint of $AC$. Then $$MR \cong \frac12 AB.$$ We also have $$ER \cong AB -\frac12AC\tag{1}\label{1}$$ and $$AN \cong NE \cong \frac12(AC-AB).\tag{2}\label{2}$$Adding \eqref{1} and \eqref{2} we conclude that $MNR$ is isosceles. Since $\measuredangle MRC = 48^\circ$, form Exterior Angle Theorem we reach our result $$\boxed{\measuredangle MNC = 24^\circ}.$$

Answer 2

Here is a brute-force answer using trigonometry. By the sine rule:

$$\frac{AC}{\sin 96^{\circ}} = \frac{BC}{\sin 48^{\circ}} = \frac{AB}{\sin 36^{\circ}}$$

Without loss of generality, let $AC = 1$. Then $BC = \frac{\sin 48^{\circ}}{\sin 96^{\circ}}, AB = \frac{\sin 36^{\circ}}{\sin 96^{\circ}}$. Now let us find the coordinates of $M$ using coordinate geometry, where $A$ is located at the origin. $B = (AB \cos 48^{\circ}, AB \sin 48^{\circ}) = \left(\frac{\sin 36^{\circ} \cos 48^{\circ}}{\sin 96^{\circ}}, \frac{\sin 36^{\circ} \sin 48^{\circ}}{\sin 96^{\circ}} \right)$. Hence $M = \left(\frac{1}{2} + \frac{\sin 36^{\circ} \cos 48^{\circ}}{2 \sin 96^{\circ}}, \frac{\sin 36^{\circ} \sin 48^{\circ}}{2\sin 96^{\circ}} \right)$. Also, $E = (1 - \frac{\sin 36^{\circ}}{\sin 96^{\circ}}, 0)$ so $N = \left(\frac{1}{2} - \frac{\sin 36^{\circ}}{2 \sin 96^{\circ}}, 0 \right)$.

Thus if $x = \angle MNC$:

$$\tan x = \frac{\sin 36^{\circ} \sin 48^{\circ}}{\sin 36^{\circ} \cos 48^{\circ}+\sin 36^{\circ}}$$ $$= \frac{\sin 48^{\circ}}{\cos 48^{\circ} + 1}$$

where by the tangent half-angle formula, we recognise this as $\tan 24^{\circ}$. Hence $\angle MNC = 24^{\circ}$.

Answer 3

Alternative solution:
I will use sine rule too, but for both triangles $\triangle ABC$ and $\triangle MNC$, and well known relation: $\sin \theta = \sin (180^{\circ} - \theta)$. But first let find $CN$!

We know $CE = AB$, so $AE = AC - AB \implies NE = \dfrac{AC-AB}{2}$. Now, we have: $$ CN=CE+NE=AB+\frac{AC-AB}{2}=\frac{AB+AC}{2}. $$ If we use sine rule for triangle $\triangle ABC$, we have: $$ \frac{AB}{\sin 36^{\circ}}=\frac{AC}{\sin 96^{\circ}}=\frac{BC}{\sin 48^{\circ}}. \qquad (1) $$ If we use sine rule for triangle $\triangle MNC$, we have: $$ \frac{\dfrac{AC+AB}{2}}{\sin(180^{\circ}-(36^{\circ}+\theta))}=\frac{\dfrac{BC}{2}}{\sin \theta}. \qquad (2) $$ From $(1)$ we have: $$ AC=\frac{BC\sin 96^{\circ}}{\sin 48^{\circ}}, \quad AB=\frac{BC\sin 36^{\circ}}{\sin 48^{\circ}}, $$ hence we can rewrite equation $(2)$ as: $$ \frac{\sin96^{\circ}+\sin 36^{\circ}}{\sin 48^{\circ} \sin(36^{\circ}+\theta)}=\frac{1}{\sin\theta} \iff (\sin96^{\circ}+\sin 36^{\circ})\sin\theta=\sin 48^{\circ} \sin(36^{\circ}+\theta), $$ or $$ (\sin96^{\circ}+\sin 36^{\circ})\sin\theta=\sin48^{\circ}\sin36^{\circ}\cos\theta+\sin48^{\circ}\cos36^{\circ}\sin\theta, $$ or $$ \tan\theta = \frac{\sin48^{\circ}\sin36^{\circ}}{\sin96^{\circ}+\sin 36^{\circ}-\sin48^{\circ}\cos36^{\circ}} \implies \boxed{\theta = 24^{\circ}}. $$